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3 years ago

8

A 100 N horizontal force is applied to a 5.0 kg box causing the box to move at a constant speed of 3.0 m/s. What is the value of

the friction force on the box? 50 N 100 N 115 N 85 N

2 answers:

Angelina_Jolie [31]3 years ago

6 0

F = ma + friction

m=5kg but speed=3.0m/s=constant so a=0

F = friction = 100 N

a_sh-v [17]3 years ago

3 0

The box to move at a constant speed so there is no acceleration. The horizontal force applied is only for overcoming the friction force. So the friction force = the applied force = 100 N.

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How does the mass of an object change the gravitational pull on two objects in a system? How does the distance between objects i

LuckyWell [14K]

Answer:The greater the mass, there greater then gravitational pull on their objects.

There greater the distance,the lower the gravitational pull between the object

Explanation:

Gravitational pull(force) is directly proportional to the products Of The masses. therefore if the mass increase,the gravitational pull(force) also increases.

Gravitational pull(force) is inversely proportional to distance.if the distance between the objects increases,the gravitational pull(force) decreases .

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Three dogs (Spot, Fido, and Steinberg) are pulling on a chew toy. The chew toy is experiencing no acceleration. Spot is pulling

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Answer:

The magnitude and direction of the force applied by Steinberg are approximately 15.192 newtons and 126.704º.

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The chew toy is at equilibrium and experimenting three forces from three distinct dogs. The Free Body Diagram depicting the system is attached below. By Newton's Laws we construct the following equations of equilibrium: (<em>Sp</em> is for Spot, <em>F</em> is for Fido and <em>St</em> is for Steinberg) All forces and angles are measured in newtons and sexagesimal degrees, respectively:

$\Sigma F_{x} = F_{F}\cdot \cos \theta_{F} + F_{St,x} = 0$ (1)

$\Sigma F_{y} = F_{F}\cdot \sin \theta_{F}-F_{Sp}+F_{St,y} = 0$ (2)

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$F_{St,x} = -F_{F}\cdot \cos \theta_{F}$

$F_{St,x} = -(20\,N)\cdot \cos 63^{\circ}$

$F_{St, x} = -9.080\,N$

$F_{St,y} = F_{Sp}-F_{F}\cdot \sin \theta_{F}$

$F_{St,y} = 30\,N-(20\,N)\cdot \sin 63^{\circ}$

$F_{St, y} = 12.180\,N$

The magnitud of the force is determined by Pythagorean Theorem:

$F_{St} = \sqrt{F_{St,x}^{2}+F_{St,y}^{2}}$

$F_{St} =\sqrt{(-9.080\,N)^{2}+(12.180\,N)^{2}}$

$F_{St} \approx 15.192\,N$

Since the direction of this force is in the 3rd Quadrant on Cartesian plane, we determine the direction of the force with respect to the eastern semiaxis:

$\theta_{St} = 180^{\circ} + \tan^{-1} \left(\frac{F_{St,y}}{F_{St,x}}\right)$

$\theta_{St} = 180^{\circ} + \tan^{-1} \left(\frac{12.180\,N}{-9.080\,N}\right)$

$\theta_{St} \approx 126.704^{\circ}$

The magnitude and direction of the force applied by Steinberg are approximately 15.192 newtons and 126.704º.

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almond37 [142]

Answer:

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Explanation:

The given values are:

Momentum,

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⇒ $48.1084=45.2\times v2$

⇒ $v2=\frac{48.1084}{45.2}$

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Firlakuza [10]

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